Edexcel S2 — Question 5 15 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApproximating Binomial to Normal Distribution
TypeExact binomial then normal approximation (same context, different n)
DifficultyModerate -0.3 This is a straightforward S2 question testing standard binomial calculations (parts a,b) and normal approximation with continuity correction (part c). All parts follow textbook procedures with no novel problem-solving required, though the multi-part structure and need to apply continuity correction make it slightly more substantial than pure recall.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial

5. In a certain school, \(32 \%\) of Year 9 pupils are left-handed. A random sample of 10 Year 9 pupils is chosen.
  1. Find the probability that none are left-handed.
  2. Find the probability that at least two are left-handed.
  3. Use a suitable approximation to find the probability of getting more than 5 but less than 15 left-handed pupils in a group of 35 randomly selected Year 9 pupils.
    Explain what adjustment is necessary when using this approximation. \section*{STATISTICS 2 (A) TEST PAPER 3 Page 2}

AnswerMarks
(a) No. of left-handed is \(B(10, 0.32)\); \(P(X = 0) = 0.68^{10} = 0.0211\)M1 M1 A1
(b) \(P(X = 1) = 10 \times 0.68^9 \times 0.32 = 0.0995\)M1 A1
\(P(X \ge 2) = 1 - 0.0211 - 0.0995 = 0.879\)M1 A1
(c) Now no. of left-handed is \(B(35, 0.32) \approx N(11.2, 7.616)\)M1 A1
\(P(5 \le X < 15) = P(5-5 < X < 14.5) = P(-2.07 < Z < 1.20)\)M1 A1 A1
\(= 0.8849 - 0.0193 = 0.866\)M1 A1
Continuity correction, going from discrete to continuous variableB1
Total: 15 marks
(a) No. of left-handed is $B(10, 0.32)$; $P(X = 0) = 0.68^{10} = 0.0211$ | M1 M1 A1 |
(b) $P(X = 1) = 10 \times 0.68^9 \times 0.32 = 0.0995$ | M1 A1 |
$P(X \ge 2) = 1 - 0.0211 - 0.0995 = 0.879$ | M1 A1 |
(c) Now no. of left-handed is $B(35, 0.32) \approx N(11.2, 7.616)$ | M1 A1 |
$P(5 \le X < 15) = P(5-5 < X < 14.5) = P(-2.07 < Z < 1.20)$ | M1 A1 A1 |
$= 0.8849 - 0.0193 = 0.866$ | M1 A1 |
Continuity correction, going from discrete to continuous variable | B1 |

**Total: 15 marks**
5. In a certain school, $32 \%$ of Year 9 pupils are left-handed. A random sample of 10 Year 9 pupils is chosen.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that none are left-handed.
\item Find the probability that at least two are left-handed.
\item Use a suitable approximation to find the probability of getting more than 5 but less than 15 left-handed pupils in a group of 35 randomly selected Year 9 pupils.\\
Explain what adjustment is necessary when using this approximation.

\section*{STATISTICS 2 (A) TEST PAPER 3 Page 2}
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2  Q5 [15]}}